03-08-2010
blank space in regex pattern using sed
why does sed 's/.* //' show the last word in a line
and
sed 's/ .*//' show the first word in a line? How is that blank space before or after the ".*" being interpreted in the regex?
i would think the first example would delete the first word and the next example would delete the second word because you have .* to match any number of characters and then a space substituted with nothing, should'nt that remove the first matched word on the line that is terminated with whitespace?
---------- Post updated at 07:07 PM ---------- Previous update was at 06:49 PM ----------
I think i get it... if you do /.* // that will grab the first thing on the line if it has no space before it, and replace it will whitespace, and then print the rest of the line,
but if you do / .*// it will ignore what is before the first space, and begin substitution after the first space removing the rest of the line by substituting white space. Correct?
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LEARN ABOUT DEBIAN
strsplit
STRSPLIT(3pub) C Programmer's Manual STRSPLIT(3pub)
NAME
strsplit - split string into words
SYNOPSIS
#include <publib.h>
int strsplit(char *src, char **words, int maxw, const char *sep);
DESCRIPTION
strsplit splits the src string into words separated by one or more of the characters in sep (or by whitespace characters, as specified by
isspace(3), if sep is the empty string). Pointers to the words are stored in successive elements in the array pointed to by words. No
more than maxw pointers are stored. The input string is modifed by replacing the separator character following a word with '